All categories

3 years ago

• what is the typical distance between two adjacent pins on a 14-pin dual-in-line ic package?

1 answer:

muminat3 years ago

7 0

A 14 pin dual-in-line IC package[14 DIL] is an integrated socket which is most popular form of IC package and has a wide range of application in digital electronics.

The 14-pin DIL has two pairs per side and each pair contains seven connecting pins.

The pairs of pins are arranged linearly one after another.The typical dimensions of width is 6.5 mm and the typical dimension of length is 18 mm.

we are asked to calculate the typical distance between two adjacent pins.

The typical distance between two adjacent pins is calculated as-

$Typical\ distance =\frac{dimensional\ length}{number\ of\ pins\ in\ each\ row}$

$=\frac{18 mm}{7}$

$= 2.5714 mm$ [ans]

You might be interested in

If themass is 50kg, what weight of water is to be displaced to float on water? why

ludmilkaskok [199]

Answer:if youre looking for the weight of the thermas in genral it should be 500n

Explanation:using the formula w=mg

w=500x10

giving us 500 newtons which is the weight.

5 0

1 year ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago

Suppose a clay model of a koala bear has a mass of 0.200 kg and slides on ice at a speed of 0.750 m/s. It runs into another clay

kaheart [24]

Answer:

0.278 m/s

Explanation:

We can answer the problem by using the law of conservation of momentum. In fact, the total momentum before the collision must be equal to the total momentum after the collision.

So we can write:

$mu=(m+M)v$

where

m = 0.200 kg is the mass of the koala bear

u = 0.750 m/s is the initial velocity of the koala bear

M = 0.350 kg is the mass of the other clay model

v is their final combined velocity

Solving the equation for v, we get

$v=\frac{mu}{m+M}=\frac{(0.200)(0.750)}{0.200+0.350}=0.278 m/s$

8 0

3 years ago

what distance is a book from the floor if the book contains 196 joules f potential energy and has a mass of 5 kg?

AleksAgata [21]

E=mgh. 196=5kg*9.81m/s^2*h. So h=196/(5*9.81)=4m

5 0

3 years ago

When conducting scientific research, any information that is found must be analyzed to determine its validity. John believes tha

SashulF [63]

Answer:

b. "One thing for sure is that they were UFOs and they had to be piloted by some type of intelligent life."

Explanation:

Well...generally the above observation may seem wrong but it isn't. That's because according to scientific point of view, if he saw a real UFO flying out in the air then it is logically possible that some being of an advance civilization must be piloting that craft. If the observation is correct then it can't just be a coincidence at all.

5 0

3 years ago

Read 2 more answers